What are the concentrations of Pb²+ and Cu²† when the cell potential falls to 0.37 V? Express your answers using two significant figures separated by a comma. ? [Pb²*], [Cu²*] = mol L-1 %3D

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What are the concentrations of Pb²+ and Cu2+ when the cell potential falls to 0.37 V?
Express your answers using two significant figures separated by a comma.
ΑΣΦ
?
[Pb**], [Cu**] =
mol L-1
%3!
Transcribed Image Text:What are the concentrations of Pb²+ and Cu2+ when the cell potential falls to 0.37 V? Express your answers using two significant figures separated by a comma. ΑΣΦ ? [Pb**], [Cu**] = mol L-1 %3!
A voltaic cell consists of a Pb/Pb²+ half-cell and a
Cu/Cu²+ half-cell at 25 °C. The initial concentrations
of Pb2+ and Cu²+ are 0.0520 mol L and 1.60
mol L1, respectively.
Pb2+ (ag) + 2 e → Pb(8)
Cu²+ (aq) + 2 e¯ → Cu(s)
E° = -0.13 V
E° = 0.34 V
Transcribed Image Text:A voltaic cell consists of a Pb/Pb²+ half-cell and a Cu/Cu²+ half-cell at 25 °C. The initial concentrations of Pb2+ and Cu²+ are 0.0520 mol L and 1.60 mol L1, respectively. Pb2+ (ag) + 2 e → Pb(8) Cu²+ (aq) + 2 e¯ → Cu(s) E° = -0.13 V E° = 0.34 V
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